The number of solutions of the equation $$\sin^{-1}\left[x^2 + \frac{1}{3}\right] + \cos^{-1}\left[x^2 - \frac{2}{3}\right] = x^2$$ for $$x \in [-1, 1]$$, and $$[x]$$ denotes the greatest integer less than or equal to $$x$$, is:

Given equation,

$$\sin^{-1}\left[x^2+\frac13\right]+\cos^{-1}\left[x^2-\frac23\right]=x^2$$

for

$$x\in[-1,1]$$

First apply domain conditions.

For

$$\sin^{-1}\left[x^2+\frac13\right]$$

we need

$$-1\le x^2+\frac13\le1$$

Since

$$x^2\ge0,$$

only upper bound matters:

$$x^2+\frac13\le1$$

$$x^2\le\frac23$$

Similarly, for

$$\cos^{-1}\left[x^2-\frac23\right],$$

we need

$$-1\le x^2-\frac23\le1$$

This is automatically satisfied for

$$x^2\in\left[0,\frac23\right]$$

Hence valid domain is

$$x^2\le\frac23$$

Now use identity

$$\sin^{-1}t+\cos^{-1}t=\frac\pi2$$

Write

$$x^2-\frac23=\left(x^2+\frac13\right)-1$$

Put

$$u=x^2+\frac13$$

Then equation becomes

$$\sin^{-1}u+\cos^{-1}(u-1)=x^2$$

Now observe that

$$u\in\left[\frac13,1\right]$$

Checking possible values:

If

$$u=1,$$

then

$$x^2=\frac23$$

LHS becomes

$$\sin^{-1}(1)+\cos^{-1}(0)$$

$$=\frac\pi2+\frac\pi2=\pi$$

which is not equal to

$$\frac23$$

For all valid

$$x,$$

LHS lies in

$$\left[\frac\pi2,\pi\right]$$

while RHS satisfies

$$x^2\le\frac23$$

Since

$$\frac\pi2>\frac23,$$

equation cannot hold.

Hence there are no real solutions.

Therefore, the required number of solutions is

$$\boxed{0}$$